Acceleration is sometimes expressed in multiples of g, where is the acceleration of an object due to the earth's gravity. In a c

Question

2 years ago

6

ar crash, the car's forward velocity may go from to in How many g's are experienced, on average, by the driver?

1 answer:

katen-ka-za [31] · Answer 1 · 2022-07-27T21:27:24+03:00

<h2>Answer:</h2>

Driver experiences 17.68 G's on an average.

<h2>Step-by-step explanation:</h2>

In the question,

We know that in a car crash the speed of the car goes from the speed of 26 m/s to 0 m/s in 0.15 seconds.

So,

The final speed, v = 0 m/s

Initial speed, u = 26 m/s

Also,

Time in seconds, t = 0.15 s

So,

Using the equation of law of the motion, we get,

v = u + at

0 = 26 + a(0.15)

a = -26/0.15

a = -173.33 m/s²

As, there is deceleration of the car is taking place the acceleration is in negative.

Now,

To convert it into the force in the G's we need to divide the deceleration by the acceleration (gravity), g.

So,

G's experienced by the driver on an average is given by,

$\frac{173.33}{9.8}=17.68$

Therefore, on an average the driver experiences 17.68 G's.